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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Differential Equations

NDA 2015 · Mathematics · Differential Equations

Consider the following statements: 1. The general solution of (d...

Consider the following statements: 1. The general solution of dydx=f(x)+x\frac{d y}{d x}=f(x)+x is of the form y=g(x)+cy=g(x)+c, where cc is an arbitrary constant. 2. The degree of (dydx)2=f(x)\left(\frac{d y}{d x}\right)^2=f(x) is 2 . Which of the above statements is/are correct?

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

C. Both 1 and 2

Step-by-step solution

Statement 1: The general solution of dydx=f(x)+x\frac{d y}{d x}=f(x)+x Now integrating on both side ∫dydx=∫[f(x)+x]dx∴y=f(x)+C∴\int \frac{d y}{d x}=\int[f(x)+x] d x\therefore \quad y=f(x)+C\therefore \quad Statement 1 is correct. Statement 2: dydx=f(x)+c\frac{d y}{d x}=f(x)+c Squaring both sides, (dydx)2=(f(x)+c)2(dydx)2=[f(x)2+2cf(x)+c2]\begin{aligned} & \left(\frac{d y}{d x}\right)^2=(f(x)+c)^2 \\ & \left(\frac{d y}{d x}\right)^2=\left[f(x)^2+2 \mathrm{cf}(x)+c^2\right] \end{aligned} Hence, the differential equation is of order -1 and degree -2 . ∴\therefore \quad both Statements 1 and 2 are correct. ∴\therefore \quad Option (C) is correct.
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